Question

f(x) = |sin x + cos x + tan x + cot x + sec x + csc x|

• A. 1
• B. 2
• C. 0
• D. 3

Answer 1

Answer: (A)

1

Answer 2

Let $u = \sin x + \cos x$. The range of $u$ is $[-\sqrt{2}, \sqrt{2}]$. We can rewrite the function as $f(x) = |u + \frac{2}{u-1}|$. Let $g(u) = u + \frac{2}{u-1}$. We need to find the minimum value of $|g(u)|$. The derivative $g'(u) = 1 - \frac{2}{(u-1)^2}$. Setting $g'(u)=0$ gives $u = 1 \pm \sqrt{2}$. The value $u = 1+\sqrt{2}$ is outside the range. The value $u = 1-\sqrt{2}$ is inside the range. At $u = 1-\sqrt{2}$, $g(u) = 1-\sqrt{2} + \frac{2}{(1-\sqrt{2})-1} = 1-\sqrt{2} + \frac{2}{-\sqrt{2}} = 1-\sqrt{2} - \sqrt{2} = 1-2\sqrt{2}$. As $u \to 1^-$, $g(u) \to -\infty$. As $u \to 1^+$, $g(u) \to +\infty$. At $u = -\sqrt{2}$, $g(-\sqrt{2}) = -\sqrt{2} + \frac{2}{-\sqrt{2}-1} = -\sqrt{2} + \frac{2(-\sqrt{2}+1)}{(-\sqrt{2})^2-1^2} = -\sqrt{2} + \frac{-2\sqrt{2}+2}{2-1} = -\sqrt{2} - 2\sqrt{2} + 2 = 2-3\sqrt{2}$. At $u = \sqrt{2}$, $g(\sqrt{2}) = \sqrt{2} + \frac{2}{\sqrt{2}-1} = \sqrt{2} + \frac{2(\sqrt{2}+1)}{(\sqrt{2})^2-1^2} = \sqrt{2} + \frac{2\sqrt{2}+2}{2-1} = \sqrt{2} + 2\sqrt{2} + 2 = 3\sqrt{2}+2$. The range of $g(u)$ for $u \in [-\sqrt{2}, \sqrt{2}]$ (excluding $u=\pm 1$) is $(-\infty, 1-2\sqrt{2}] \cup [3\sqrt{2}+2, \infty)$. The minimum value of $|g(u)|$ is $|1-2\sqrt{2}| = 2\sqrt{2}-1 \approx 1.828$. The least integer less than or equal to $2\sqrt{2}-1$ is $1$.